Groupy Theory: Cosets

It should be common knowledge now that I have trouble with Group Theory. I would like to go back and start from the beginning but I haven't the luxury of time at this point. So for the present time I am resigned to just keeping up with the class the best I can. For anyone has the time and patience, I would appreciate it if someone can look over my work for the following 2 questions. Advice on how to approach the question, hints, interpretations of concepts, and expansions on concepts are welcome.

ii) Calculate the product AB = (Hy)(Hy2) of the cosets Hy and Hy2 (ie., write down and simplify every possible product ab, where a is an element of A and b is an element of B).

iii) Is AB a coset of H in G?

iv) Is AB a coset of any subgroup of G? (Hint: Use Lagrange's Theorem).

i)

A = Hy = {uy, xy} = {y, xy}
B = Hy2 = {uy2, xy2} = {y2, xy2}.

ii)

y*y2 = y3 = u
y*xy2 = xy2

xy*y2 = xy3= xu = x
xy*xy2 = x*xy-1*y2 = u*y = y

Therefore AB = {u, y, x, xy2}

iii)

I am not sure about this part of question 1, but I would think that AB is not a coset of H in G since AB has 4 elements while H has only two.

iv)

I am also not sure about this part of question 1. However, I think that since the order of AB is 4 and that the order of G is 6, 4 is not a divisor of 6 hence AB cannot be a subgroup of G. If there cannot be a subgroup of order 4, AB cannot be a coset of any subgroup since there are no subgroup of order 4. (?)

Question2:

i) Let G be a group, and let H be a subgroup of G. What condition tells you that H is a normal subgroup of G?

ii) Prove the following: H is normal in G iff g-1Hg = H for every g which is an element of G.

i)

A subgroup H of a group G is a normal subgroup of G if the following is true:

Condition: gH = Hg for every g which is an element of G. That is, the right coset Hg of H in G, generated by g, is equal to the left coset gH of H in G, generated by g (where g is an element of G).

Question 1 Part iii:
Your reasoning is correct, but this depends on part ii.

Question 1 Part iv:
Your reasoning is correct but the results from Part ii have changed

Question 2 Part i:
Looks good

Question 2 Part iii

What you have is probably acceptable. It's unnecessary to introduce x, since you already have g-1.

So you have, for example
H=g-1Hg (by hypothesis)
then multiply both sides by g on the left
g*H=g*g-1Hg
gH=uHg
gH=Hg

If you want to be more formal, H,Hg, gH and g-1Hg are all sets, so you can consider operating on the elements in the sets, but there's not a whole lot of extra insight to be gained:
e.g
by hypothesis we have that
H=g-1Hg
then for every
h in H there is h' in g-1Hg such that
h=g-1h'g
multiply both sides by g on the left
gh=gg-1h'g
so
gh=h'g
This is true for every h in H, so gH is a subset of Hg.
Hg and gH also have the same number of elements.
Therefore Hg=gH